module 4: answer key - open school...

Module 4: Answer Key

Section 1: Logic/Proof

Lesson 1 Inductive and Deductive Reasoning 167

Lesson 2 AND, OR, and NOT and Venn Diagrams 171

Lesson 3 Counterexamples, Quantifiers, and Negations 175

Lesson 4 Conditional Statements 177

Lesson 5 Direct and Indirect Proofs 181

Review 185

Section 2: Functions

Lesson 1 Performing Operations on Functions 191

Lesson 2 Inverse Functions 197

Lesson 3 Remainder Theorem and Factor Theorem 203

Lesson 4 Polynomial Functions and Their Graphs 211

Review 219

Section 3: Functions on the Graphing Calculator

Lesson 1 Polynomial Functions With a Graphing Calculator 229

Lesson 2 Graphs of Rational Functions 235

Review 243

Principles of Mathematics 11 Answer Key, Contents 165

Module 4

166 Answer Key, Contents Principles of Mathematics 11

Module 4

Module 4

Lesson 1

Answer Key

1. a) The rule is that all Senior Years students must takeEnglish. Tracy is a student so she will take it. This isdeductive reasoning.

b) In the past, tests have always been the day after a hockeygame. Based on this, a generalization is made. This isinductive reasoning.

c) The general rule is that anyone who likes to playbadminton likes to play tennis. Because Jeff likesbadminton, he must like tennis. This is deductivereasoning.

d) The rule is that all sides of an equilateral triangle arecongruent. More specifically, ∆ ABC is equilateral andAB = AC. This is deductive reasoning.

e) Based on the 20 minute of observation, she drew hergeneralization. This is inductive reasoning.

2. (a), (c), and (d) are valid. While the others may be likely, youlack sufficient evidence to draw a valid conclusion.

3. ∠ + ∠ =

= =

∴ =

α β

α β

α β

90o

By definitionBCAB

BCAB

sin cos

sin cos

β

AC

B

α

Principles of Mathematics 11 Section 1, Answer Key, Lesson 1 167

4. Each point on the circumference may be joined to (n – 1)other points to form (n – 1) chords. Similarly, there are(n – 1) chords from every other starting point. Hence, thereare n(n – 1) chords altogether. But, each chord is actuallycounted twice; according to its beginning and its end. And so,

the number of distinguishable chords is

5. It appears that the number of regions follows the pattern2

1, 2

2, 2

3, 2

4. . . 2

n. But draw the diagram for 6 points and

count up the number of regions.

Note that inductive reasoning only gives you a conjecturethat you can test further. In this case, the test for 6 pointstells you that your conjecture was not correct.

n n − 12

.

168 Section 1, Answer Key, Lesson 1 Principles of Mathematics 11

Module 4

( )

Points Regions

2 2

3 4

4 8

5 16

Look for the small triangle in the middle somewhere.

6. Tangents GE and GF are congruent, because tangents drawnto a circle from an external point are equal. Because amedian is a line drawn from a vertex of a triangle to themidpoint of the opposite side, FD = ED. is the commonside so ∆ GFD ≅ ∆ GED by SSS and ∠ 1 and ∠ 2 arecorresponding angles of congruent triangles.

GD

1 7

13

14

12

1 1

26

16

9

22

20

1 8

2

1030

23

8

15

2 824 19

2 7 3

42 5

57

61

31

29

2 1


Module 4


Module 4

Notes

Module 4

Lesson 2

Answer Key

1. a) A and B

b) A or B

c) not A and not B

d) not (A or B)

A B

A B

A B

A B


2. a) X and Y = {1, 2, 3} or Yb) X or Z = {1, 2, 3, 4, 5, 6, 8}c) Y ∩ Z = φ (no elements in common)d) Y ∪ Z = {1, 2, 3, 4, 6, 8}e) = {6, 7, 8}f) = {6, 7, 8}g) not X and not Y = {6, 7, 8}h) = {4, 5, 6, 7, 8}

3.

4.

Let U = total number of studentsB = students liking bluegrassC = students liking country and western

(Note: Alternatively, you can find B ∩ C by solving theequation 21 – x + x + 14 – x + 18 = 46.)

Because seven like both bluegrass and country and western,then 21 – 7 or 14 like bluegrass but not country and western.

not (B or C)B or C

but B or C B CB C

B C

== − == + −= + −= + −=

1846 18 2821 14

28 21 1421 14 287

∩∩

∩

B C

14−x

18

21−x x

U

D E

F

+o +

o

o

+

o

x

xo

+

o

X Y∩

X Y∪X


5. T = tennisS = soccerR = rugby6 goes in the overlap ofthe 3 circles13 play tennis andsoccer so 7 completesT ∩ S11 play rugby andsoccer so 5 completesR ∩ S36 play tennis and rugby so 30 completes T ∩ RNumber for tennis only = 163 – 30 – 6 – 7 = 120.R ∪ T = 208R ∪ T = R + T – R ∩ T208 = R + 163 – 36R = 208 – 163 + 36 = 81Those that only play rugby = 81 – 30 – 6 – 5 = 40R ∪ S = 98R ∪ S = R + S – R ∩ S98 = 81 + S – 1128 = SSoccer only = 28 – 18 = 10Total membership = sum of the numbers in each region =218

T R

S

1 2 0 30

76

40

1 0

5


Module 4

6. a)

b)

c)

d)

e)

f)

5 4 3 2 1 0 1 2 3 4 5 6

5 4 3 2 1 0 1 2 3 4 5 6

5 4 3 2 1 0 1 2 3 4 5 6

5 4 3 2 1 0 1 2 3 4 5 6

Empty Set

5 4 3 2 1 0 1 2 3 4 5 6

5 4 3 2 1 0 1 2 3 4 5 6


Module 4

Module 4

Lesson 3

Answer Key

1.

2. Since 2 > 1, the conjecture is false.

3.

Since the denominator of f(x) cannot be equal to zero, x 7.

4. a) All squares are parallelograms.b) Some parallelograms are rectangles.

c) No octagon has nine sides.

d) Some triangles are isosceles.

5. a) All apples are green.b) All students like mathematics.

c) It is raining.

d) At least one triangle has two right angles.

6. a) someb) no

c) some

d) all

∴ = −−

≠ +when xxx

x7497

72

,

xx

x xx

x2 49

77 7

77

−−

=+ −

−= +

If thenx = =12

112

2, .

00


is not defined

≠

( )( )


Module 4

Notes

Lesson 4

Answer Key

1. a) If I go to the theatre, then it is Saturday.b) If the ceiling fan is running, then the light switch is on.c) If Jim does his laundry, then it is Monday night.d) If ice cream is left on the counter, then it will melt.e) If Marie takes the bus, then she is late for her job

interview.

2. 2x + 3 = 13 if and only if x = 5.

3. a) Converse: If an apple is green, then it is not ripe.Statement is false; converse is false.

b) Converse: If you have a car, then you are popular.Statement is false, converse is false.

c) Converse: If the angles opposite two sides of a triangleare congruent, then the two sides of the triangle arecongruent.Statement is true; converse is true.

4. a) Contrapositive: If an apple is not green, than it is ripe.Statement is false, contrapositive is false.

b) Contrapositive: If you do not have a car, you are notpopular.Statement is false; contrapositive is false.

c) Contrapositive: If the angles opposite two sides of atriangle are not congruent, then the two sides are notcongruent.Statement is true; contrapositive is true.

5. a) Inverse: If an apple is ripe, then it is not green.Statement is false, inverse is false.

b) Inverse: If you are not popular, then you do not have acar.Statement is false; inverse is false.

c) Inverse: If you do not have the measles, then you do nothave spots.Statement is true; inverse is false.


Module 4

6. a) Statement: If Melanie lives in Kamloops, then Melanielives in BC. — TrueConverse: If Melanie lives in BC, then Melanie lives inKamloops. — FalseInverse: If Melanie does not live in Kamloops, thenMelanie does not live in BC. — FalseContrapositive: If Melanie does not live in BC, thenMelanie does not live in Kamloops. — True

b) Statement: If May 4 is a Monday, then May 5 is aTuesday. — TrueConverse: If May 5 is a Tuesday, then May 4 is a Monday.— TrueInverse: If May 4 is not a Monday, then May 5 is not aTuesday. — TrueContrapositive: If May 5 is not a Tuesday, then May 4 isnot a Monday. — True

c) Statement: If Tartan has four legs, then Tartan is a dog.— FalseConverse: If Tartan is a dog, then Tartan has four legs. — TrueInverse: If Tartan does not have four legs, then Tartan isnot a dog. — TrueContrapositive: If Tartan is not a dog, then Tartan doesnot have four legs. — False

d) Statement: If it is snow, then it can be melted. — TrueConverse: If it can be melted, then it is snow. — FalseInverse: If it is not snow, then it can not be melted. —FalseContrapositive: If it can not be melted, then it is notsnow. — True

7. a) If a quadrilateral does not have congruent diagonals,then it is not a rectangle.

b) If you do not have to take a driving test, then you have alicence.


Module 4

8. (One possible answer) If x is an even number, then x isdivisible by 2.

9. (One possible answer) If ∆ ABC is equilateral, then ∆ ABCis isosceles.

10. Converse: If p + q is a multiple of 11, then p and q are bothmultiples of 11. — False

Contrapositive: If p + q is not a multiple of 11, then p and qare not multiples of 11. — True

To show the converse is false, let p = 8 and q = 3. p + q is amultiple of 11 but p and q are not multiples of 11.


Module 4


Module 4

Notes

Module 4

Lesson 5

Answer Key

1. a) My sister likes to play ringette.b) In ∆ ABC, ∠ A = ∠ B.

2. a) indirect proofb) direct proof, ∠ 1 = ∠ 3c) indirect proof

3. a) The kitchen cutlery is not stainless steel.b) The kitchen cutlery is stainless steel.c) The kitchen cutlery should not rust.d) The kitchen cutlery is beginning to rust.

4. a) K is blue so M is red and J is blue.b) Because J is not blue, M is not red. Because M is not red,

K is not blue.

5. Given: Scalene ∆ ABF

You are to prove thatthe bisector of anyangle F is notperpendicular to .Assume that thebisector of F isperpendicular to .

∠ AQF = ∠ BQF = 90º

FQ = FQ

∠ AFQ = ∠ BFQ since FQ bisects ∠ F

Then ∆ AFQ ≅ ∆ BFQ by ASA and AF = BF.

This assumption that leads to a contradiction thatscalene ∆ ABF is isosceles.

∴ The bisector of any angle of a scalene triangle cannot beperpendicular to the opposite side.

FQ AB⊥

AB

AB

F

AQ

B


6. There are three possibilities for the tallest person: Velda,Surrinder, or Harpreet. If Velda is the tallest, (a) and( c) areboth true. But we should have only one true statement — sonot Velda. If Harpreet is the tallest, statements (b) and (c)would both be true. Again, too many true ones. If Surrinderis the tallest, (a) and (c) are both false and so (b) will be thetrue statement if Harpreet is the next in height and Velda isthe shortest. And so, Surrinder is the tallest, Harpreet isnext, and Velda is the shortest.

7. First use an indirect proof to show that A is a knight.a) Make the assumption: A is not a Knight. It follows that A

is a Knave. Then the statement he made (“At least one ofus is a Knave.”) must be false since Knaves always lie.Therefore, both men, in particular A, would be Knight,which contradicts the assumption that A is not a Knight.Thus, the assumption that A is not a Knight must befalse.

It follows that A must be a Knight and so his statement istrue. Because one of the pair is a Knave and A is aKnight, B must be a Knave.

b) Suppose C were a Knight. Then his statement would betrue, which would mean he is a Knave. This contradictionshows that C cannot be a Knight, so C is a Knave. But ifC is a Knave, he has to lie and say he is a Knight. No oneon this island can say “I am a Knave”. This statement iscalled a paradox and cannot be resolved in the context ofthis problem. It is an indication that either the data orthe initial assumptions are flawed.


Module 4

8.

Assume is a median. Then by the definition ofa median.

BDA = CDA = 90º is an altitude

is common

ABD ACD SAS

so corresponding sides

This contradicts the fact that so is not amedian.

9. Prove that the diagonals of a trapezoid do not bisect eachother. Assume that the diagonals do bisect each other withAO = OC and DO = OB.

If the diagonals of a quadrilateral bisect each other, thequadrilateral is a parallelogram. If ABCD is a parallelogram,then ABCD has two pairs of opposite sides parallel. Butbecause it is a trapezoid, it has exactly one pair of parallelsides. This is contradictory. Our assumption is false. Thediagonals do not bisect each other.

10. In this one, you divided by 0 in the form (b – a) according tothe first line.

ADAB AC≠

AB AC=

AD

AD

BD CD=AD

≠Given AB AC

AD is an altitude

Prove AD is not a median.


Module 4

A

CBD

∠∠

∆ ∆≅


Module 4

Notes

Review

Answer Key

1.

2.

a = number who read The Globe and Mail and The National Post

b = number who read The National Post and Timec = number who read The Globe and Mail and Time

88 = 6 + 42 + a + c 40 = a + c (1)76 = 5 + 42 + a + b 29 = a + b (2)85 = 8 + 42 + b + c 35 = b + c (3)Eqn. (1) – Eqn. (2): 11 = c – b

35 = c + bAdd Eqn. (1) and Eqn. (2): 46 = 2c

23 = c12 = b17 = a

NG

T

5a

6

42

c b

8

7688

85x

U

113

a) A B (Xs)

b) C B (Os)

c) Not A (*)

Principles of Mathematics 11 Section 1, Answer Key, Review 185

Module 4

O

X

X

OO O

O

O

*

A B

C

*

**

∩

∪

x can be found by adding the numbers within the three circlesand subtracting the sum from 113

x = 113 – (6 + 42 + 17 + 23 + 5 + 12 + 8) = 0

This means every one of the business executives reads at leastone of the three publications.

3.

4. The diagonals of all quadrilaterals bisect each other.

5. a) Converse: If it can be melted, then it is snow.Validity — False

Inverse: If it is not snow, then it cannot be melted.

Validity — False

Contrapositive: If it cannot be melted, it is not snow.

Validity — True

b) Converse: If the angles opposite two sides of a triangle arecongruent, then the two sides are congruent.

Validity — True

Inverse: If the two sides of a triangle are not congruent,then the angles opposite those sides are not congruent.Validity — True

Contrapositive: If two angles of a triangle are notcongruent, then the sides opposite the angles are notcongruent.

Validity — True

in a trapezoid or other non-parallelogram, the diagonals do

not intersect at their midpoints

28 – 5 – 11 – 3 = 9students who did notreceive a B in eithermathematics orEnglish5 11 3

2 8

M E

186 Section 1, Answer Key, Review Principles of Mathematics 11

Module 4

6. Ben is the murderer. Indirect proof.

7.

Since PA and PB are tangents from P, PA = PB and ∠ PAB= ∠ PBA. The angle at P = 60°. Since the angles at A and Bequal 120°, then all three angles are equal to 60° and allthree sides are congruent. The triangle is, therefore,equilateral.

8. a) Converse

b) If the sum of the exterior angles of a polygon is 360°, thenthe polygon is convex.

c) If two quadratic functions do not have the same axis ofsymmetry, they do not have the same vertex.

9. Given sets A, B, and C as shown in the Venn diagram below,shade A or B (A ∪ B).

5

31

4

2 2 3

A

B C

P

A

B

60°


Module 4

10.Given A and B shown in the Venn diagram, find A or B.

Answer: 20

11.State whether the following statements are examples ofdeductive or inductive reasoning.

All isosceles triangles have two congruent sides.

∆ABC is isosceles, therefore, it has 2 congruent sides.

Answer: deductive

12.A total of 32 students taking Senior 3 Pre-Calculus areinvolved in volleyball, tennis, and basketball. It was foundthat:

19 play volleyball

25 play volleyball or basketball

11 play volleyball and tennis

4 play volleyball, tennis, and basketball

8 play basketball and tennis

7 play basketball and volleyball

Complete the Venn diagram and use it to find the number ofstudents who take tennis.

A B*

***

***

** *

*

** ** ** *

* *


Module 4

Tennis: 22

V o l l e y b a l l B a s k e t b a l l

T e n n i s

5 3

47 4

7

2


Module 4


Module 4

Notes

Module 4

Lesson 1

Answer Key

1. a) f(2) = 6. You are finding the value of f(x) when x = 2.

b) f(3) = 7. You are finding the value of f(x) when x = 3.

c) g(6) = 10. You are finding the value of g(x) when x = 6.d) g(7) = 12. You are finding the value of g(x) when x = 7.

e) g(f(2)) = g(6) = 10. Find f(2), then find g(6).

f) g(f(3)) = g(7) = 12. Find f(3), then find g(7).

g) g(f(4)) = g(7) = 12. Find f(4), then find g(7).

h) f(2) + g(7) = 6 + 12 = 18. Add the two values.i) f(3) – g(6) = 7 – 10 = –3. Subtract the two values.

j) f(4) • g(7) = (7)(12) = 84. Multiply the two values.

k) Divide the two values.

l) f(g(6)) = f(10) = undefinedFind g(6) and then find f(10). There is no ordered pair in fthat has an x-value of 10.

2. f(x) = 3x + 4 and g(x) = x2 – 1

a)

b)

c)

d) f g f g− = − = − − =

= ⋅ + =

1 0 1 1 1 0

3 0 4 4

2

g f g f− = − = − + =

= − =

1 1 1 3 1 4 1

1 1 02

f g f g2 3 2 2 1 3

3 3 4 13

2= = − =

= ⋅ + =

g f g f2 10 2 3 2 4 10

10 1 992

= = ⋅ + =

= − or

f

g

27

612

12

05= = or . .


( ) ( )( )

( ) ( )( )

( ) ( )( ) ( )

( )

( )

( ) ( )( ) ( ) ( )

( )

( ) ( )

e)

f)

g)

h)

3.

a)

b)

c) t s t s8 7 812

8 3 7

2 7 68

= = + =

= ⋅ −=

s t s t− = − − = − − = −

= − +

= −

3 12 3 6 6 12

12

12 3

3

s t s t7 8 7 2 7 6 8

12

8 3 7

= = ⋅ − =

= ⋅ + =

t x x s x x= − = +2 612

3and

g g a g a g a a

a

a a

a a

= − = −

= − −

= − + −

= −

2 2

2 2

4 2

4 2

1 1

1 1

2 1 1

2

f f a f a f a a a

a

a a

= + = ⋅ + = +

= + +

= + + +

3 4 3 4 3 4

3 3 4 4

9 12 4 9 16or

f g a f a g a a

a

a a

= − = −

= − +

= − + +

2 2

2

2 2

1 1

3 1 4

3 3 4 3 1or

g f a g a f a a a

a

a a

a a

= + = + = +

= + −

= + + −= + +

3 4 3 4 3 4

3 4 1

9 24 16 1

9 24 15

2

2

2


Module 4

( ) ( )( ) ( ) ( )

( ) ( )( ) ( )

( ) ( )( ) ( )

( ) ( )( ) ( )( )

( )

( ) ( )

( ) ( ) ( )

( )( )

( )

( )

( ) ( ) ( ) ( )

( )

( )

( )

( )

( )

d)

e)

f)

Yes, s(t(x) = t(s(x). This will be discussed in the next lesson.

4. a)

i) h

g

6 6 2 4 2

9 2 9 18

= − = =

= ⋅ =

g h

h g

g

h

6

9

2

18

2 218 2416

44

1

=

= ⋅−

= =or

h x x g x x= − =2 2and

t s a t a s a a

a

a a

= + = +

= + −

= + − =

12

312

3

212

3 6

6 6

s t a s a t a a a

a

a a

= − = ⋅ − = −

= − +

= − + =

2 6 2 6 2 6

12

2 6 3

3 3

t s t s− = − = − + =

= ⋅ −

= − = −

34

218

34

12

34

3218

2218

6

214

634


Module 4

( )( ) ( )

( )

( )

( )( ) ( )

( )( )

( )( )

( ) ( )

( ) ( )

a +12

3

a +12

3

s218

− 34

− 34

− 34

ii)

iii)

iv)

b) i) Domain: {x|x 1} or [1, )

Range: {y|y 0} or [0, )

ii) Domain: {x|x 2} or [2, )Range: {y|y 0} or [0, )

5. a)

=−

− =− =

= ++= ≠

1To find the range solve for

1( 1) 1

111

so 0

y xx

y x

xy y

xy y

yx y

y

f k x f x D x x

f k xx x

R y y

= + = ≠ −∞ ∞

=+ − −

= ≠ −∞ ∞

1 1 1 1

11 2

11

0 0 0

{ | } , ,

{ | } , ,

or

or or

g h x h x

x

= −

= −

2

2 2

− ≥ ≥

−

2 2 0 so 1

2 2 implies the positive root

x x

x

h g x h x

x

=

= −

2

2 2

Rationalize

Simplify 8 2 2=

h

g

5 5 2 3

5 2 5 10

= − =

= ⋅ =g h

h g

g

h

5

5

3

10

2 310 2

2 38

2 32 2

32

62

=

= ⋅−

=

=

=

=


Module 4

( )( )

( ) ( ) ( )

( )

( ) ( )

( ) ( )∪

∪

∞∞

∞∞

≥≥

≥≥

( )( ) ( )

( )( ) ( )

( )

( ) ( )

( )

( ) ( ) ( )

( )

6. f(x) = x3

f(5) = 53

or 125 units3

f(5) represents the volume of the cube when the edge is5 units


Module 4


Module 4

Notes

Module 4

Lesson 2Answer Key

1. a) The inverse of dividing x by 2 is multiplying x by 2.

b) The inverse of subtracting 8 from x is adding 8 to x.

c)

d)

e)

Write y = f(x)

Interchange x and y

Simplify

f x x

y x

x y

x y

x y x

= +

= +

= +

− =

− = ≥

1

1

1

1

1 12

,

( )

( )−

= + ≥

= + ≥= + ≥

− = ≥

− =

= − ≥

2

2

2

2

1

1, 0

1, 0

1, 0

1 , 0

1

1 1

f x x x

y x x

x y y

x y y

x y

f x x x

Write y = f(x)

Interchange x and y

Simplify

Replace y by f –1(x)

f xx

yx

xy

x y

x y

f x x

=+

= +

= +

= +− =

= −−

23

23

23

3 23 2

3 21

f x x f x x= − = +−8 81

f xx

f x x= ∴ =−

221


( ) ( )

( )( )

( )

( )

( )

( )

2. These functions are inverses if f(g(x)) = x and g(f(x)) = x.

a)

Because f(g(x)) = g(f(x)) = x, f(x) and g(x) are inverses.

b)


c)


g x x

g f x g x

x

x

x

= −

= +

= + −

= + −=

3

3

33

8

8

8 8

8 8

f x x

f g x f x

x

x

x

= +

= −

= − +

==

8

8

8 8

3

3

33

33

g x x

g f x g x

x

x

=

=

==

3

3

33

f x x

f g x f x

x

x

=

=

=

=

3

3

33

g xx

g f x g x

x

x

x

= +

= −

=− +

=

=

322 3

2 3 32

22

f x x

f g x fx

x

x

x

= −

= +

= + −

= + −=

2 3

32

23

23

3 3


Module 4

∴

∴

∴

( )( )

( )( )

( )

( )( ) ( )( )

( ) ( )

( )( )( )( ) ( )( )

( ) ( )

( )( ) ( )( )

( ) ( )

x + 32

x + 32

( )

d)

∴ Because f(g(x)) = g(f(x)) = x, f(x) and g(x) are inverses.

( )

( )( )

=−

+ =

=−

−

=+ −

=

= ×

=

322 3

32 3 2

32 3 2

33

33

g xx

xg f x g

x

x xx x

x xx

xx

x

( )

( )( )

( )

( )( )

+=

= − + − =

−−

+− −=

−+ −

−=

−

−=

−−

= •−

=

2 3

32

32 3

23

23 26

2 23

26 3 6

23

23

23

223

2 3

xf x

x

f g x fx

x

xx

x x

xx

x

xx

x

xxx

x

x


Module 4

3. a)

b)

c)

Let y = f(x)

Interchange x and y-values

Simplify

Replace y by f –1(x)

f xx

x

yx

x

xy

y

x y y

xy x y

xy y x

y x x

yx

x

f xx

xx

=−

=−

=−

− =− =

− =− =

=−

=−

≠−

3

3

3

3

33

1 3

31

31

11

x ≥ 0 and y ≥ 2Let y = f(x)


Simplify and solve for x(you have to add therestriction on x)Replace y by f

–1(x)

f x x

y x

x y

x y

x y x

f x x x

= +

= +

= +

− =

− = ≥

= − ≥−

2

2

2

2

2 2

2 2

2

1 2

,

,

Let y = f(x)


Simplify and solve for y

Replace y by f–1

(x)

( )

( )

( )

( )−

=+

=+

=+

+ =+ =

− = −− = −

−=−

−= ≠−

1

2

2

2

2

22

1 2

21

21

1

xf x

xx

yx

yx

y

x y y

xy x y

xy y x

y x x

xy

xx

f x xx


Module 4

( )

( )

( )

( )

( )

( ) ( )

( )

d) f xx

x

yx

x

xy

yxy y

xy y

y x

yx

f xx

x

= −

= −

= −

= −− = −− = −

= −−

= −−

≠−

3

3

3

33

1 3

31

31

11 ,


Module 4

( )

( )

( )


Module 4

Notes

Module 4

Lesson 3Answer Key

1. a)

b)

c)

d) Answer:

x x2 2 2 3 1− + −2 3 2

3

2

2

2 30 2 2 3 4 7

2 0 4

3 0 73 0 6

1

xx x x x x

x x

xx

++ − + − −

+ −

+ −+ −

−

Answer:

3 2 2 4 32x x x− − +3 2 6 16 17 62 4 3

6 4

12 17

12 89 69 6

0

3 2

2

3 2

2

2

x x x xx x

x x

x x

x x

x

x

− − + −− +

−− +− +

−−

Answer:

x x+ − +2 3 22x x x xx

x x

x

x

+ + − −−

+

− −− −

2 2 3 43

2

3 43 6

2

3 2

2

3 2

Answer:

x x x− + + +2 2 4 5 112x x xx x

x x

x x

x x

x

x

− + − ++ +

−−−

+−

2 2 0 3 12 4 5

2 4

4 3

4 85 15 10

11

3

2

3 2

2

2


( ) ( )

( ) ( )

( ) ( )

( ) ( )

e)

2. a)

b)

c)

d) − ++

2Quotient: 2 4 4Remainder: 5

x x12

2 5 6 3

1 2 2

2 4 4 5

−

−

−

+ + ++

3 2

Quotient:

3 15 5Remainder: 45x x x

5 1 2 0 70 20

5 15 75 25

1 3 15 5 45

− −

+ + ++

3 2Quotient:3 6 12 23Remainder: 42

x x x2 3 0 0 1 4

6 12 24 46

3 6 12 23 42

− −

Quotient: x2 + 2x – 3

Remainder: none

2 1 0 7 6

2 4 6

1 2 3 0

−

−

−

Answer:

x x x x+ − − +1 3 4 123 2x x x x xx x x

x x

x x

x x

x x

x x

x

x

+ − − + +− − +

+− −− −

− +− −

++

1 2 7 8 123 4 12

3 7

3 3

4 8

4 412 1212 12

0

4 3 2

3 2

4 3

3 2

3 2

2

2


Module 4

( ) ( )

e)

3. You can do this in one of three ways:

a) Synthetic division

Because the remainder is 0, then x + 2 is a factor.

b) Long division


c) If f(–2) = 0, then x + 2 is a factor


f x x x

f

= + −

− = − + − −

= − + −=

3 2

3 2

4 8

2 2 4 2 8

8 16 8

0

x x xx x

x x

x

x x

x

x

+ + + −+ −

+++− −− −

2 4 0 82 4

2

2 0

2 44 84 8

0

3 2

2

3 2

2

2

− −

− − +

− ←

2 1 4 0 8

2 4 8

1 2 4 0 remainder

Quotient:Remainder: none

4 2 42x x− −− − −

−

− −

32

4 4 7 6

6 3 6

4 2 4 0


Module 4

( )

( ) ( ) ( )

4. a) f(x) = x3

– 8x2

+ 4x + 48

Possible a values

Note: Since the leading coefficient is 1, then its factorsare ±1.

To find a remainder, find where f(a) = 0.

Use synthetic division to find a second factor:

x2 – 4x – 12 is the other factor, which factors into (x – 6)(x + 2).

The factors are (x – 4)(x2

– 4x – 12) or (x – 4)(x – 6)(x + 2).

4 1 8 4 48

4 16 48

1 4 12 0

−

− −

− −

f

x

4 4 8 4 4 4 48

64 128 16 48

0 4

3 2= − + +

= − + += ∴ − is a factor.

f

x

3 3 8 3 4 3 48

27 72 12 48

15 3

3 2= − + +

= − + += ∴ − is not a factor.

f

x

2 2 8 2 4 2 48

8 32 8 48

32 2

3 2= − + +

= − + += ∴ − is not a factor.

f x x x x

f

x

= − + +

= − + +

= − + += ∴ −

3 2

3 2

8 4 48

1 1 8 1 4 1 48

1 8 4 48

45 1 is not a factor.

=

= ± ± ± ± ± ± ±± ± ±

factors of constant termfactors of leading coefficient

1 2 3 4 6 8 1216 24 48, , , , , , ,

, ,


Module 4

( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

∴

∴

b) f(x) = x4

– 2x3

– 17x2

+ 18x + 72

Possible a values

Use synthetic division to find another factor:

x3 – 4x2 – 9x + 36 can still factor

f

x

3 3 4 3 9 3 36

27 36 27 36

0 3

3 2= − − ⋅ += − − += ∴ − is a factor.

− − −

− −

− −

2 1 2 17 18 72

2 8 18 72

1 4 9 36 0

( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

24 3

4 3 2

24 3

4 3 2

1 1 2 1 17 1 18 1 721 2 17 18 7272 1 is not a factor.

1 1 2 1 17 1 18 1 721 2 17 18 7240 1is not a factor.

2 2 2 2 17 2 18 2 7216 16 68 36 7240 2 is not a factor.

2 2 2 2 17 2

f

x

f

x

f

x

f

= − ⋅ − + ⋅ += − − + += ∴ −

− = − − − − − + − += + − − += ∴ +

= − ⋅ − + += − − + += ∴ −

− = − − − − − + ( )18 2 7216 16 68 36 720 2 is a factor.x

− += + − − += ∴ +

=

= ± ± ± ± ± ± ±± ± ± ±


1 2 3 6 8 9 1218 24 36 72

, , , , , , ,, , ,


Module 4

( ) ( )

Use synthetic division again: use x = 3 and x3 – 4x2– 9x + 36.

The quotient is x2 – x – 12 which factors into (x – 4)(x + 3).

The factors are (x + 2) (x – 3) (x + 3) (x – 4).

c) f(x) = x3 – 27

Possible “a” factors: ±1, ±3, ±9, ±27

Do synthetic division:

Other factor is x2 + 3x + 9 — does not factor further.

Factors are (x – 3) (x2 + 3x + 9).

d) f(x) = 3(4x3 – 4x2 + x – 1)

Possible a values

f

x

1 3 4 1 4 1 1 1

3 4 4 1 1

0 1

3 2= − + −

= − + −= ∴ − is a factor.

=

= ±± ± ±

= ± ± ±


11 2 4

112

14, ,

, ,

3 1 0 0 27

3 9 27

1 3 9 0

−

f

x

3 3 27

27 27 0 3

3= −= − = ∴ − is a factor.

3 1 4 9 36

3 3 36

1 1 12 0

− −

− −

− −


Module 4

( ) ( ) ( ) ( )

∴

∴

∴

( )

Use synthetic division to get another factor.

The three factors are 3(4x2 + 1)(x – 1).

5.

6. The volume of the rectangular prism is V = 3x

3+ 8x

2– 45x – 50.

Use synthetic division with V and x + 1.

The quotient is 3x2

+ 5x – 50 which factors into (3x – 10)(x + 5).The missing dimension is 3x – 10.

− − −

− − +

−

1 3 8 45 50

3 5 50

3 5 50 0

f x x x x

x x x x x

x x x

= + − + +

= − + + − + += − + +

2 1 4 3

2 2 8 4 3

2 7 7

2

3 2 2

3 2

1 4 4 1 1

4 0 1

4 0 1 0

− −


Module 4

( ) ( ) ( )

∴

∴

∴

7.

∴ = + − +p x x x x4 2 3 113 2

Eqn (2) x 2

4 2 141

4 2 142 2 2

6 122

12 1

3

b c

b c

b c

b c

bb

b c

c

c

− =+ = −

− =+ = −

==

+ = −+ = −

= −

p b c

b c

b c

p b c

b c

b c

b c

− = − + − + − + −− + − + = −

− =

= + + +

−

2 4 2 2 2 11 7

32 4 2 11 7

4 2 14 1

1 4 1 1 1 11

4 1115 14

1 2

3 2

3 2

,

,

which equals

which equals 14

14 = + + ++ + =

+ =


Module 4

( )

( )

( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

Module 4

Lesson 4

Answer Key

1. It is a graph that you can sketch without raising your pencilfrom the paper.

2. The graph of f(x) = |x| is a V-shaped curve so the turn issharp and not smooth.

3. It is an even degree polynomial and the leading coefficient ispositive, so it opens up. It rises to the right and the left.

4. The maximum number of turns is (n – 1) where n is thedegree of the polynomial function.

a) Because the degree, n, is 3, then there are (3 – 1) or 2turns.

b) Because n = 6, there are 6 – 1 or 5 turns possible.

c) Because n = 2, there is 2 – 1 or 1 turn possible.

d) Because n = 5, there are 5 – 1 or 4 turns possible.

e) Because n = 4, there are 4 – 1 or 3 turns possible.

5. a) Because the degree is odd and the leading coefficient isnegative, the graph rises to the left and falls to the rightor falls from the second quadrant.

b) Because the degree is even and the leading coefficient ispositive, the graph rises to the left and right or opens up.

c) You can determine that the degree is 3 if you multiply thethree factors together. Because the degree is odd but theleading coefficient is positive, the graph rises to the rightbut falls to the left or rises from the third quadrant.

d) Because the degree is even and the leading coefficient isnegative, the graph falls to the left and right or opensdown.

e) Because the degree is odd and the leading coefficient isnegative, the graph falls to the right and rises to the leftor falls from the second quadrant.

f) Because the degree is odd and the leading coefficient ispositive, the graph rises to the right and falls to the left orrises from the third quadrant.


6. a) — v)b) — iii)c) — ii)d) — vi)e) — i)f) — vii)g) — iv)h) — viii)

7. a) i)

ii) (n – 1) turns where n = 3 is 3 – 1 or two turns at most

iii) Because the degree of the function is odd and the leadingcoefficient is positive, the graph falls to the left and risesto the right or rises from the third quadrant.

iv)

Using sign analysis you can determine that in theinterval

• x < –2, the curve is below the x-axis• –2 < x < 0, above the x-axis• 0 < x < 2, below the x-axis• x > 2, above the x-axis.

f x x x x

f x x x x

f x

x x x

x x x

= − −

= − +

=

∴ − + =

= = = −

3 2 6

3 2

0

3 2 0

0 3 2

Common factor

Let

or, ,


Module 4

y

x

A •

• • •

( )

( ) ( )

( ) ( )

( ) ( )

Choose points between –2 and 0 to find the approximatepeak (y = 4) and between 0 and 3 to find the approximatevalue (y = –7.875) or use your graphing calculator and thecalc menu - see Section 3, Lesson 1, if you don’t know howto do this on your calculator.

b) i)

ii) Number of turns (n – 1) = (3 – 1) = 2 turns at most

iii) Because the degree of the function is odd and the leadingcoefficient is negative, the graph falls to the right andrises to the left or falls from the second quadrant.

iv)

f x x x

f x x x

f x x x x

f x

x x x

x x x

b gb g d ib g b gb gb gb gb g

= − +

= − −

= − − +

=

− − + =

= = = −

3

2

4

4

2 2

0

2 2 0

0 2 2

Let to find the zeros

or , ,


Module 4

x

A

Bf(x) = –x3 + 4x

y

c) i) f(x) = x4

– 2x3

– 3x2

+ 4x + 4Use the remainder and factor theorem to find the factors.

Possible zeros are the factors of 4: ± 1, ± 2, ± 4

If f(–1) = 0, then x + 1 is a factor and –1 is a zero.

Use synthetic division to get a new quotient to factor:

New quotient: x3 – 3x2 + 4

x + 1 is a factor and x = –1 is also a root of x3 – 3x2 + 4 so x = – 1 is a double root.

Use synthetic division to get a new quotient:

New quotient is x2 – 4x + 4 which factors as (x – 2)2.

(x – 2)2 (x + 1)2 = 0

(x – 2)2 = 0 (x + 1)2 = 0

x = 2 x = –1Double zeros of 2 and –1.

ii) Number of turns (n – 1) or (4 – 1) or three turns at most.

iii) Because the degree of the function is even and theleading coefficient is positive, the graph rises to the rightand to the left or opens up.

− − +

− −

−

1 1 3 0 4

1 4 4

1 4 4 0

f x x x

f

= − +

− = − − − +

= − − +=

3 2

3 2

3 4

1 1 3 1 4

1 3 4

0

− − − + +

− −

−

1 1 2 3 4 4

1 3 0 4

1 3 0 4 0

f x x x x x

f

= − − + +

− = − − − − − + − +

= + − − + =

4 3 2

4 3 2

2 3 4 4

1 1 2 1 3 1 4 1 4

1 2 3 4 4 0


Module 4

∴

∴

( )

( )

( )

( ) ( ) ( ) ( )

( ) ( ) ( )

iv)

d) i) Zeros

ii) Number of turns: (n – 1) or 4 turns, at most.iii) Because the degree of the function is odd and the leading

coefficient is negative, the graph rises to the left and fallsto the right or falls from the second quadrant.

iv) You can use a reflection of y = x5 over the x-axis and avertical transformation of f(x) = –x5 shifted two unitsdownward.

on your calculator use(–2)^(0.2) or (–2)^(115)

Let f x

x

x

x

=

− − == −≈ −

0

2 0

2115

5

5

.


Module 4

x

y5 •

• •–1 2

( )

x

y

e) i)

ii) Number of turns is 3 – 1 or 2 turns, at most.

iii) Because the degree of the function is odd and the leadingcoefficient is positive, the graph of the function rises tothe right and falls to the left or rises from the thirdquadrant.

iv)

This is a horizontal transformation of y = x3 shifted4 units to the right.

f) i) f(x) = x3

+ 5x2

+ 2x – 8Possible zeros are the factors of 8: ± 1, ± 2, ± 4, ± 8

If f(a) = 0, then x – a is a factor and a is a zero.

x – 1 is a factor and x = 1 is a zero.

Use synthetic division to get a new quotient:

f 1 1 5 1 2 1 8

1 5 2 8

0

3 2= + + −= + + −=

( )( )

=

− =

− == =

3

Let 0

4 0

4 04 4 is a triple root

f x

x

x

x x


Module 4

∴

( ) ( ) ( )

x

y

4

x2 + 6x + 8 is the new quotient which factors into(x + 4)(x + 2) = 0

(x – 1)(x + 4)(x + 2) = 0

zero = – 2, – 4, 1

ii) Number of turns is 3 – 1 or 2 turns, at most.

iii) Because the degree of the function is odd and the leadingcoefficient is positive, the graph of the function rises tothe right and falls to the left or rises from the thirdquadrant.

iv)

1 1 5 2 8

1 6 8

1 6 8 0

−


Module 4

A

•• •

B•

•

x

y

∴

∴

8. Volume = height x length x width

= x(12 – 2x)(10 – 2x)

Use your graphing calculator and the trace function to findthe maximum volume. The value of x is ≈ 1.81 cm and themaximum volume is 96.770564 cm

3.

Hint: Try a window of: Xmin = 0Xmax = 6Xscl = 1Ymin = 0Ymax = 120Yscl = 20

12

10 xx

xx

xx

xx

10 2x

12 2x


Module 4

Review

Answer Key

1. f(x) = 2x – 1 g(x) = 5x2 + 1

a) b)

c) d)

2. f(x) = x2 + 1 g(x) = 2x – 3

a) b) f g x f x

x

x x

x x

= −

= − +

= − + += − +

2 3

2 3 1

4 12 9 1

4 12 10

2

2

2

g f x g x

x

x

x

= +

= + −

= + −

= −

2

2

2

2

1

2 1 3

2 2 3

2 1

f g

g

f f

3 3

3 5 3 1

46

3 46 49

2 49 198 197

2

+

= ⋅ +=

+ == ⋅ −= −=

f f x f x

x

x

x

= −

= − −

= − −= −

2 1

2 2 1 1

4 2 1

4 3

f g x f x

x

x

x

= +

= + −

= + −

= +

5 1

2 5 1 1

10 2 1

10 1

2

2

2

2

g f x g x

x

x x

x x

x x

= −

= − +

= − + +

= − + += − +

2 1

5 2 1 1

5 4 4 1 1

20 20 5 1

20 20 6

2

2

2

2


Module 4

( )( ) ( )( )

( )( )

( )( ) ( )( )

( )

( ) ( )( )

( )

( )( )( )

( )

( )

( )( )

( )( )( )

3. b)

c)

4. a) g(x) = 2(x – 1)2 + 2

x

y

l

( )

( )

2

2

1 2

3, 3, 0

3

3

3

3

3, 0, 3

f x x x y

y x

x y

x y

x y

f x x x y−

= − ≥ ≥

= −

= −

= −

+ =

= + ≥ ≥

( )

( )

( )1

3, 2

23

23

2

2 3

2 32 3

, 0, 2

2 3

f x xx

yx

xy

x y

xy xx

y x yx

xf x

x−

= ≠−

=−

=−

− =

− =+

= ≠ ≠

+=

( ) 2

2

2

2

1, 0

1

1, 0

1

1

1, 1, 0

f x x x

y x

x y y

y x

y x

y x x y

= + ≥

= +

= + ≥

= −

= ± −

= − ≥ ≥


Module 4

b) x = 2(y – 1)2

+ 2

5.

a)

b)

c)

f

f

f aa

a

−

−

−

= − = − = −

= − =

+ = + − =

1

1

1

11 7

36

32

88 7

313

3 73 7 7

3

f x x

y x

x y

xy

f xx

= += += +

− =

= −−

3 7

3 73 7

73

73

1

x

y


Module 4

( )

( )

( )

( )

( )

6. If f(x) and g(x) are inverses of each other

Since they both equal x, they are inverse functions.

7. f(x) = x3 + 2x2 – 5x – 6

a) If x + 1 is a factor, then f(–1) = 0

x + 1 is a factor of f(x)

b) If x – 3 is a factor of f(x), then f(3) = 0

f(3) 0, then x – 3 is not a factor.

f 3 3 2 3 5 3 6

27 18 15 624

3 2= + ⋅ − −= + − −=

f − = − + − − − −= − + + −=

1 1 2 1 5 1 6

1 2 5 6

0

3 2

( )( ) ( )( )( ) ( )1

2 12

1 2 1 12 12 2

21 12

f g x g f x x

xf g x

x x

xx

x x

= =

− = +

− + − + =

− + =

=


Module 4

( ) ( ) ( ) ( )

( ) ( )

∴ ≠

∴

8. a) f(x) = x3

– 2x2

+ 3x – 6Possible a values = ±1, ±2, ±3, ±6

If f(1) = 0, then x – 1 is a factor

x – 1 is not a factor.

If f(2) = 0, then x – 2 is a factor

x – 2 is a factor.

f(x) = (x – 2)(x2 + 3)

b) f(x) = 2x3

+ 3x2

– 32x + 15Possible values or zeros:

∴ = − + −

= − − +

f x x x x

x x x

3 2 9 5

3 2 1 5

2

3 2 3 32 15

6 27 15

2 9 5 0

−

−

−

f x

f

x

f x

1 2 1 3 1 32 1 15 12 1

1 2 1 3 1 32 1 15 48

1

3 2 3 3 3 32 3 15 0 3

2

3 2

3 2

= ⋅ + ⋅ − + = − ∴ −

− = − + ⋅ − − − + =∴ +

= + − + = ∴ −

,

,

,

is not a factor

is not a factor

is a factor

± ± ± ± ± ± ± ±12

132

3 552

15152

, , , , , , ,

2 1 2 3 6

2 0 6

1 0 3 0

− + −

f 2 2 2 2 3 2 6

8 8 6 6

0

3 2= − + −= − + −=

f 1 1 2 1 3 1 6

1 2 3 64

3 2= − ⋅ + −= − + −= −


Module 4

∴

∴

∴

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )( )( )( ))(

∴∴

9. Since the remainder is –2x + 1 when ax4

+ bx3

– 8x2

+ 6 isdivided by x

2– 1, subtract –2x + 1 from the polynomial to

form f(x) = ax4

+ bx3

+ 8x2

+ 2x + 5.

a + b = 1

a – b = 5

Solving: a = 3, b = –2

10. a) f(x) = x(x – 1)(x + 3)

zeros: 0, 1, –3

y-intercept = 0

x

y

13

+

0

+

f a b

f a b

1 1 1 8 1 2 1 5 0

1 1 1 8 1 2 1 5 0

4 3 2

4 3 2

= + − =

− = − + − − − + − + =

Since

then

x x x

f

f

2 1 1 1

1 0

1 0

− = − +

=

− =


Module 4

Note: since

p(x) = (x2 – 1)(q(x)) +r(x)

p(x) – r(x) = (x2 –1)q(x)

let

f(x) = p(x) –r(x)

− −

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

∴

∴

( )

( )

b) f(x) = (x – 1)2(x + 4)

2

zeros: 1, –4

y-intercept = 16

c) f(x) = x3 + 4x2 + x – 6

possible zeros: ±1, ±2, ±3, ±6

If f(1) = 0, then f is a zero.

f(1) = 13 + 4 • 12 + 1 – 6

= 0

zeros at 1, –2, –3

f x x x x

x x x

= − + +

= − + +

1 5 6

1 3 2

2

1 1 4 1 6

1 5 6

1 5 6 0

−

x

y

16

40

14

+++


Module 4

−

( ) ( )( )∴

( )( )( )

y-intercept = –6

x

y

l l l

l

1− 3

− − +

0

+

− 2


Module 4

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Module 4

Lesson 1

Answer Key

1. a) i)

Using sign analysis you can determine that in theinterval

• x < –2, the curve is below the x-axis • –2 < x < 0, above the x-axis• 0 < x < 2, below the x-axis• x > 2, above the x-axis.

Choose points between –2 and 0 to find the approximatepeak (y = 4) and between 0 and 3 to find the approximatevalue (y = –8.2088) or use your graphing calculator andthe calc menu - see Section 3, Lesson 1, if you don’t knowhow to do this on your calculator.

ii) No absolute maximum or minimum values. However,there is a local maximum at A where y = 4 and at Bwhere y ≈ –8.2088.


Module 4

x

A

f(x) = x3

– x2

– 6x

lll

l

y

b) i)

ii) There is no absolute maximum or minimum. There is alocal maximum at A where y = 3 and a local minimum atB where y = –3.

c) i)

ii) Absolute minimum when x = –1 and y = 0 and at x = 2when y = 0. Relative maximum when x = 0.51 and y = 5.1.


Module 4

x

A

Bf(x) = –x

3+ 4x

y

l

l

ll

l

x2

4

–1

y

ll

l

d) i) You can use a reflection of y = x5

over the x-axis and avertical transformation of f(x) = –x

5shifted two units

downward.

ii) There is no maximum or minimum.

e) i)

This is a horizontal transformation of y = x3 shifted4 units to the right.

ii) There is no maximum or minimum value.


Module 4

x4

y = (x – 4)2

y

x

y

f) i)

ii) There is no absolute maximum or minimum values. A isa local maximum achieved at y ≈ 4.1 when x is –3 and Bis a local minimum achieved when x ≈ –0.21 and y =–8.2.

2. Two points to remember about the Graphing Calculatorgraphs in this answer key:

• You were asked to hand-sketch your graphs, but we havereproduced the actual screen from a TI-83 calculator. As longas your hand-sketched graphs have the same general shapeas graphs here, give yourself full marks.

• Your [xmin, xmax] and [ymin, ymax] values are probablydifferent for most graphs. As a result, your sketches maylook off-center, stretched, or compressed compared to thesegraphs. This is to be expected, and not a cause for concernunless your graph was so badly off center that its minimaand maxima are hard to discern.

a) solve x3 – x2 – 12x = –3

Y1 = X^3 - X^2 - 12X + 3

Solution set:{-3.13, 0.25, 3.89}

x [-8, 8]; [-20, 20]


Module 4

x

A

B

y

l

ll

l

l

y = x3

+5x2

+ 2x – 8

b) solve –x3 + 2x2 – x + 1 = 0

Y1 = -X^3 + 2X^2 - X + 1

Solution set:{1.75}

x [-2, 3]; [-2, 2]

3. Give the relative maximum and relative minimum forx3 – 3x2 = 9x – 9

Y1 = X^3 - 3X^2 - 9X + 9

Relative Max: (-1, 14)Relative Min: (3, -18)

x [-5, 7]; [-20, 20]

4. Find the smallest zero of the function y = 2x4 + 5x3 – x2 + 5x – 11

Y1 = 2X^4 + 5X^3 - X^2 +5X - 11

Solution set: one zeronear -3 and one near +1which has the lowerabsolute value. The latteris at exactly (1,0).

x [-5, 5]; [-35, 20]


Module 4

5. Determine the local maximum and minimum points of thegraph of the function y = x(x – 5)(x + 3)

Y1 = X(X-5)(X+3)

Relative Max: (-1.67, 14.8)

Relative Min: (3, -36)

x [-5, 8]; [-45, 20]

6. Determine the Range of the function y = –x4 – 3x3 + 3x2 + 11x + 6

Y1 = -X^4 - 3X^3 + 3X^2 - 11X + 6

To two decimal places, the upperrelative max is: (1.47, 16.31). Nopoint is higher on the y-axis thanthat, so the range is:y ∈ [-∞, 16.31]

x [-5, 5]; [-15, 20]


Module 4

Module 4

Lesson 2Answer Key

1. a) i) x-intercept is the zero of the numeratorx – 1 = 0, x = 1

ii) y-intercept is the value of

iii) Equation of vertical asymptote are vertical linesthrough the zero of the denominator:

Equation of horizontal asymptote: Because thedegree of the numerator is less than the degree of thedenominator, the x-axis is the horizontal asymptotewhich has the equation y = 0.

b) i) x-intercept is the zero of the numerator; that is, whenx = 0

ii) y-intercept is the value of

iii) Equation of vertical asymptote are vertical linesdrawn through the zero of the denominator

Equation of horizontal asymptote: Because thedegree of the numerator is less than the degree of thedenominator, the x-axis is the horizontal asymptote.Its equation is y = 0.

x xx x

− = + == = −

3 0 3 03 3

f 00

0 90=

−=

x x

x xx x

− − =− = − == =

2 3 0

2 0 3 02 3

( ) ( )( ) ( ) ( )0 1 1 1

00 2 0 3 2 3 6

f− −

= = = −− − − −


( )

( )( )

c) i) x-intercept: x2

– 4 = 0, x = ±2ii) y-intercept

iii) Vertical asymptotes: Because the denominator x2 + 4has no real zeros, there are no vertical asymptotes. Horizontal asymptotes: Because the degree of thenumerator is equal to the degree of the denominator,y = the ratio of the leading coefficient of numerator tothat of the denominator. Therefore, y = 1.

2. a) — ii)b) — i)c) — iv)d) — iii)

3. a)

There are no factors common to the numerator or to thedenominator.x-intercept: None (no variable in numerator)

y-intercept:

Vertical asymptote: x – 4 = 0, x = 4Horizontal asymptote: y = 0Sign analysis: x > 4 – curve is below the x-axis

x < 4 + curve is above the x-axis

f 01

0 414

=−−

=

f xx

= −−14

x

y

vertical

asymptotex = 4

horizontalasymptotey = 0

y -intercept

f 00 40 4

1=−+

= −


Module 4

( )

( )

( )

b)

There are no factors common to the numerator or to thedenominator.x-intercept: 3x = 0, x = 0

y-intercept:

Vertical asymptote: x + 2 = 0, x = –2Horizontal asymptote: since the degrees are equal, the

horizontal asymptote equation is

Choose points or do a sign analysis to complete the graph.

c) f xxx

= +−

23

x

y

y-intercept

y = 1

x = 3

x -intercept

y y= =31

3or

f 03 00 2

0=⋅+

=

f xx

x=

+3

2

x

y

x- and y-intercepts

y = 3

x = 2


Module 4

( )

( )

( )

There are no factors common to the numerator or to thedenominator.x-intercept: Let numerator = 0x + 2 = 0, x = –2

y-intercept:

Vertical asymptote: x – 3 = 0, x = 3Horizontal asymptote: — since degrees of numerator and

denominator are equal

Using selected ordered pairs or sign analysis, completethe graph.

d)

There are no factors common to the numerator or to thedenominator.x-intercept: Since the numerator has no variable, there isno x-intercept.

y-intercept:

Vertical asymptote: (x – 3)(x + 3) = 0x = 3 or x = –3

Horizontal asymptote: y = 0, because the degree of thenumerator is less than degree of the denominatorChoose points that work in each region or use signanalysis.

Find f 03

0 913

=−

= −

f xx

=−3

92

x

y

y -intercept

y = 0

x = 3

x = 3

y = =11

1

Find f 00 20 3

23

= +−

= −


Module 4

( )

( )

( )

e)

There are no factors common to the numerator or to thedenominator.x-intercept: Let numerator = 0 to find its zeros, which arethe x-intercept of the functions (x – 2)(x + 2) = 0x = 2 or –2

y-intercept:

vertical asymptote: (x – 3) = 0, (x + 3) = 0x = –3 or x = 3

Horizontal asymptote: y = 1, because the numerator anddenominator have equal degrees.Choose ordered pairs in each region or do a sign analysis.

f)f x

xx xx x

x x

xx

f xxx

x

= −+ −

=− ++ −

= ++

= ++

≠

2

2

12

1 1

2 1

1212

1,x

y

y-intercept

y = 1

x = 2

x-intercept

Find f 00 40 9

49

=−−

=

f xxx

= −−

2

2

49

x

y

y-intercept

y = 1

x = 3

x = 3

x -interceptx-intercept


Module 4

( )

( )

( )

( )

( )( )( )( )

There are common factors so you exclude x = 1 from thedomain.x-intercept: Let numerator = 0 x + 1 = 0, x = –1

y-intercept:

Vertical asymptote: x + 2 = 0 x = –2Horizontal asymptote: Degrees of numerator anddenominator are equal.

g)

There are no factors common to numerator ordenominator.x-intercept: Let x

2= 0, x = 0

y-intercept:

Vertical asymptote: (x – 1) (x + 1) = 0x – 1 = 0, (x + 1) = 0x = 1 or x = –1

Horizontal asymptote: y = 1Use ordered pairs to complete graph or do a sign analysis.

Find f 00

0 10=

−=

f xx

x=

−

2

2 1

x

y

y = 1

x = 1x = 1

∴ = =y11

1

Find f 00 10 2

12

=++

=


Module 4

( )

( )

( )

→

∴

4. a)

The numerator and denominator have common factors.x-intercept: noney-intercept: Find f(0) = –1 Vertical asymptote: x – 1 = 0, x = 1Horizontal asymptote: y = 0Complete graph.

b)

The numerator and denominator have no common factors.x-intercept: x = 0y-intercept: f(0) = 0 Vertical asymptote: x2 – 1 = 0

(x – 1)(x + 1) = 0x – 1 = 0 x + 1 = 0x + 1 x = – 1

Horizontal asymptote: y = 0Choose ordered pairs to complete the graph and/or do asign analysis.

f xx

x=

−2 1

x

y

y = 0

x = 1x = 1

x- and y-intercepts

f xxx

xx x

f xx

x

= +−

= +− +

=−

≠ −

11

11 1

11

1

2

x

y

y = 0

x = 1

y-intercept


Module 4

( )

( )

( )( )

( )

c)

The numerator and denominator have no commonfactors.x-intercept: 3x = 0, x = 0y-intercept: f(0) = 0 Vertical asymptote: x2 – 9 = 0

(x – 3)(x + 3) = 0x – 3 = 0 x + 3 = 0x + 3 x = – 3

Horizontal asymptote: y = 0Complete graph.

d)

The numerator and denominator have common factors.x-intercept: noney-intercept: f(0) = 2 Vertical asymptote: noneHorizontal asymptote: y = 0Complete graph.

f xx

=+

212

x

y

f xx

x=

−3

92

x

y

y = 0

x = 3x = 3


Module 4

( )

∴

( )

Review

Answer Key

1. a)

V.A.: x = 1

H.A.: y = 2

x-intercept: –2

y-intercept: –4

x

y

y = 2

x = 1

1

++

2

f xxx

= +−

2 41


Module 4

( )

−

b)

V.A.: x = 2 or x = –2H.A.: y = 0

x-intercept: none

y-intercept:

c)

V.A.: x – 3 = 0 or x + 3 = 0

H.A.: y = 0

x-intercept: 0

y-intercept: 0 3

++

3 0

f xx

x=

−2 9

x

y

− 12

2

++

2

f xx

=−2

42


Module 4

−

−

−−

−

( )

( )

2. x3

– 5x + 2 = 0

Y1 = x3 – 5x + 2

Y2 =

Y3 =

Y4 =

x [–10, 10] y [–10, 10]

The zeros of this function give the roots of the equation. Therefore the solutions are –2.41, 0.41, 2.00

3. a)

Y1 = x4 – 6x3 + 5x2 + 4x – 6

Y2 =

Y3 =

Y4 =

x [-5, 8] y [–50, 50]

x

y


Module 4

The solution to x4

+ 5x2

= 6x3– 4x + 6 is where the

graph of Y1 is below the x-axis.

Therefore the solutions are: –0.92, 4.85

b) y ≥ –39.35 since the minimum value is –39.35

4.

5.

3 24 8 14 R –27x x x+ − + =

–2 1 6 0 – 2 1

– 2 –8 16 –281 4 – 8 14 –27


Module 4

•

•

•

•

• (0.4, 7.8)

(2.6, 1.3)

(–3, –1)

xy = 3

y = 8 – x2

x

y

module 4: answer key - open school...

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